Explanation Given: The arbitrary velocity of body is x ˙ 0 . Concept used: Write the equation of motion a system containing spring, mass and damper. m x ¨ + c x ˙ + k x = 0 ...... (1) Here, m is the mass, k is the spring constant, c is the damping constant, x ˙ is the velocity of damper and x ¨ is the acceleration of mass. For heavy damping the solution for equation (1) is. x = c 1 e λ 1 t + c 2 e λ 2 t ...... (2) Here, x is the displacement of system, λ 1 and λ 2 are the roots of equation (1) and t is time. The above solution corresponds to a non-vibratory motion, as 1 and 2 are both negative, x approaches to zero as t increases indefinitely. Differentiate equation (2) with respect to time t . x ˙ = c 1 λ 1 e λ 1 t + c 2 λ 2 e λ 2 t ...... (3) Here, x ˙ is the velocity. Calculation: Substitute x 0 for x and 0 for t in equation (2). x 0 = c 1 + c 2 ...... (4) Here, x 0 is any arbitrary position. Substitute x ˙ 0 for x ˙ and 0 for t in equation (3). x ˙ 0 = c 1 λ 1 + c 2 λ 2 ...... (5) Substitute x 0 − c 1 for c 2 in equation (5) and solve for c 1 . c 1 = − x ˙ 0 + λ 2 x 0 λ 2 − λ 1 Substitute − x ˙ 0 + λ 2 x 0 λ 2 − λ 1 for c 1 in equation (4)

12th Edition

ISBN: 9781259977305

Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.

Publisher: Mcgraw-hill Education,

See similar textbooks

Concept explainers

Free Damped Vibrations

Free vibration is a type of vibration in which the external force is removed after giving the initial displacement to a body/system. In other words, in this type of vibration, force is applied at the initial displacement of the system, and after the init…

Question

Chapter 19.5, Problem 19.128P

To determine

To show for heavy damping a body released from an arbitrary position with an initial velocity cannot pass more than once through its equilibrium position.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 19.5, Problem 19.127P

Chapter 19.5, Problem 19.129P

Students have asked these similar questions

A force of 20 newton stretches a spring 1 meter. A 5 kg mass is attached to the spring, and the system is then immersed in a medium that offers a damping force numerically equal to 10 times the instantaneous velocity. (a) Let x denote the downward displacement of the mass from its equilibrium position. [Note that x>0 when the mass is below the equilibrium position. ] Assume the mass is initially released from rest from a point 3 meters above the equilibrium position. Write the differential equation and the initial conditions for the function x(t)(b) Solve the initial value problem that you wrote above.(c) Find the exact time at which the mass passes through the equilibrium position for the first time heading downward. (Do not approximate.)(d) Find the exact time at which the mass reaches the lowest position. The "lowest position" means the largest value of x

A spring with a 8-kg mass and a damping constant 2 can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons. Suppose the spring is stretched 1 meters beyond its natural length and then released with zero velocity. In the notation of the text, what is the value c2 – 4mk? m'kg/sec? help (numbers) Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t with the general form cieat cos(Bt) + cze"s t sin(&t) help (numbers) a = B = * help (numbers) * help (numbers) 8 = * help (numbers) C1 = help (numbers) C2 = 2 help (numbers) kies help us deliver our services. By using our services, you agree to our use of cookies. OK Learn more CONNECT OFF/ON CAPS Charge Po- 14 F4 F5 1- 1+ F6 F7 F8 F9 F10 F11 F12 7 8 R T Y U 5

4) Figure below shows a spring-mass- damper system. The mass is subjected to a steady force of P 2 N. The spring constant k 200 N/m, the mass m 5.2 kg and the damping ratio is f= 12.2 N. sec/m. What is the position of the mass after a sufficiently long time of force application? You may assume that the mass is initially at rest, and the spring at time t = 0 is in its unstretched condition. m k

Chapter 19 Solutions

Vector Mechanics For Engineers

Ch. 19.1 - A particle moves in simple harmonic motion....Ch. 19.1 - A particle moves in simple harmonic motion....Ch. 19.1 - Prob. 19.3P Ch. 19.1 - Prob. 19.4P Ch. 19.1 - Prob. 19.5P Ch. 19.1 - A 20-lb block is initially held so that the...Ch. 19.1 - Prob. 19.7P Ch. 19.1 - A simple pendulum consisting of a bob attached to...Ch. 19.1 - Prob. 19.9P Ch. 19.1 - A 5-kg fragile glass vase is surrounded by packing...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

To show for heavy damping a body released from an arbitrary position with an initial velocity cannot pass more than once through its equilibrium position.

Solution Summary: The author explains that a body released from an arbitrary position with an initial velocity cannot pass more than once through its equilibrium position.

Concept explainers

To show for heavy damping a body released from an arbitrary position with an initial velocity cannot pass more than once through its equilibrium position.

Want to see the full answer?

Chapter 19 Solutions